7 Ways To Support Your Child’s Maths Learning

123As parents, you can help your child want to learn maths in various ways apart from hiring maths tutors. That desire to learn will be the key to your child’s success. Enjoyment is also a crucial motivator for learning. In addition, you can point out how fortunate they are to have the great opportunity to learn maths today.

Having a good grasp of maths can open the doors to plenty of exciting possibilities.

7 Ways To Help Your Child Learn Maths

1. Offer insights into the different ways to approach maths

There are important things that your child must understand so they can develop more confidence in their maths ability.

Firstly, let them know that problems can be solved in many different ways. Learning maths goes beyond finding the correct answer. It is also a method of solving different problems and applying what they’ve learnt to new problems.

Secondly, point out that wrong answers are useful. Incorrect answers can be used to help them figure out where and why they made a mistake. Their explanation can help you discover if they need help with the concepts related to answering the problem or with number skills like division, subtraction, addition, and multiplication.

Lastly, help your child become a risk taker. Let them see the value of attempting to solve difficult problems. Allow them to explore various approaches and encourage them to speak up about their insights. This will strengthen their maths skills.

2. Promote a positive attitude towards maths

Maths can be difficult. I was not good in maths when I was a student. I did not like maths either. These statements can undermine your child’s attitude towards the subject. These comments will also give false impressions that maths is something that they can either be good at or not.

As parents, you must become a positive force in helping them learn maths. You need to let them know that solving maths problems can be satisfying, that knowledge of maths concepts is generally crucial in life, that anyone can be good at it, and that it will open up the doors to excellent career options.

3. Illustrate how maths works in day-to-day life

Your home is the best place to start exploring maths with your child. Integrate maths language and activities into daily routines to show them how this subject works in their daily life.

Sorting and matching activities will introduce your child to different mathematical operations like measurement and classification. For instance, let them sort the laundry to be washed. Ask them to put all the whites together, all the colored garments, and all the towels. As they sort things, let them count aloud how many shirts or towels are there. It is also helpful to give them the wrong number so they can count the items one by one and show to you that you have made a mistake.

Let them recognize that numbers are all around them. This will help them understand that numbers are important and that they can be used for various purposes.

4. Practice maths every day

Even when you are not at home, you can teach your child some maths concepts. In the grocery store, for instance, let them compare the prices of multipacks of vegetables and decide which of these packs offer the best value. Also, give them the opportunity to manage money by giving them some pocket money and encouraging them to budget how much will be spent. When you are on a trip, take note of the distance and the speed, and let them estimate how much time is left for your trip.

5. Let your child teach you maths

Instead of telling or showing your child how to add or multiply, it is better for you to let them teach you how they’ve learned to add or to multiply in school. Whenever you do not understand a part of the approach, let them know and ask for more clarifications. Each time they try to teach you something, they will definitely learn from that.

6. Communicate with the maths teacher

When you are concerned about your child’s learning in maths or unsure about a certain approach used in school, discuss this with their teacher.

Most educators appreciate receiving feedback, and when your child is having a hard time understanding these concepts, it is possible that other students are stuck as well.

7. Play games that buoy up mathematical thinking

Playing with blocks can teach basic maths skills like counting, recognizing symmetry, sorting, identifying patterns, and number recognition. Moreover, games with number cards will help your child come up with tactics for using numbers in several combinations by subtracting, dividing, multiplying and adding.

Helping your child to learn maths does not necessarily mean that both of you cannot have a good time and laugh. In fact, you can make games out of any maths concepts and skills. Use these activities to strengthen their maths skills as well as to build strong positive attitudes toward maths.

AUTHOR BIO
Bushra Manna is one of the founders and Principal of Leaps and Bounds Education Centre – Motorcity. She has 20 years experience teaching the British and American curricula internationally at primary level. Bushra has a passion for teaching and started her teaching career as an assistant teacher for 2 years, during which an autistic boy was appointed to her care within a mainstream classroom setting. Working with Ismail opened her eyes to the significance of knowing a child’s best learning style and having an individualized approach to teaching and building a child’s self confidence.

Using Arithmetic Signs

One of our regular visitor Aisharya Rana contributed this puzzle, which I found interesting. Hence I’m posting here to be pondered upon by all. Post your answers as comments below –

Arithmetic Puzzle

In this puzzle you need to insert any arithmetic sign in between the same digit (from 1 to 9) repeated thrice. The final result should be 6 in each case. I’m doing one for illustration –

2      2      2 = 6

can be expressed as –

2 + 2 + 2 = 6

Solve the following yourself –

1      1      1 = 6
2      2      2 = 6
3      3      3 = 6
4      4      4 = 6
5      5      5 = 6
6      6      6 = 6
7      7      7 = 6
8      8      8 = 6
9      9      9 = 6

Remember, each equation can be solved in more than one way. Post your answers in the comment section below.

Interesting Applications of Remainder Theorem

Remainder Theorem & its application

We have all learnt the Remainder Theorem in class 10 (now i am in 11) that when you divide a polynomial f(x) by x-c the remainder r will be f(c). Now let’s see how we can use this theorem in other situations.

Example #1

Let’s consider the following Product: 65 x 32.

We want to find out what is the remainder when it is divided by a number say 7.

To solve such questions we just need find the individual remainders when the numbers are divided by the divisor.

In this case 65 gives remainder 2 (65 -63) and 32 gives remainder 4 (32 – 28) when divide by 7. Multiplying the remainders we get 2*4=8

Since this number is greater than divisor, divide it again by the divisor again, i.e. 8/7 gives remainder 1.

Thus, when 65*32 is divided by 7 it gives remainder of 1. Isn’t it amazing! We save time and effort of multiplying large numbers and doing complex divisions.

Example #2

Let’s see another example to find the remainder when 1421 * 1423 * 1425 is divided by 12

By this method 1421 * 1423 * 1425

1st step remainders = 5 * 7* 9 = 35*9
2nd step remainders = 11*9
3rd step remainder = 99/12 = 3

So the monstrous product gives a remainder of 3 when divided by 12.

Example #3

Let’s suppose we want to find the last two digits of the product
22 * 31 * 44 * 27 * 37 * 43

For such problems we just need to find the remainder when it is divided by 100

(22 * 31) * (44 * 27) * (37 * 43)

1st step remainders =  82*88*91
2nd step remainder =  2 * 28

THATS IT!! The last two digits of the lengthy product is found within seconds and as you see it is 56

This is a guest post by one of the regular QuickerMaths.com follower Debasis Basak. On behalf of all the readers, I  thank him for his contribution.

The Mysterious Number 22

Number 22 Everywhere?

Numbers never fail to surprise us. This post talks about one such amazing property of number 22.

Select any three-digit number with all digits different from one another. Write all possible two-digit numbers that can be formed from the three-digits selected earlier. Then divide their sum by the sum of the digits in the original three-digit number.

You’ll always get the same answer, 22. Isn’t this wonderful!

For example, take the three-digit number 786. The 2 digit-numbers which can be made using the digits 7, 8 and 6 are 78, 87, 76, 67, 86, 68. Hence sum = 78 + 87 + 76 + 67 + 86 + 68 = 462. Sum of digits of 786 = 7+8+6 = 21. Then 462/21 = 22

This will be true for any three-digit number with all digits different.

Is it actually Mysterious?

Not really! If we go deeper and try to analyze this unusual result using, we’ll be able to appreciate the logic of it.

The general representation of any three digit number with all digits different will be 100x+10y+z.  Now to find the sum of all the two-digit numbers taken from the three digits

= (10x+y)+ (10y+x)+(10x+z)+(10z+x)+(10y+z)+(10z+y)
= 20(x+y+z) + 2(x+y+z)
=22(x+y+z)

This when divide by the sum of the digits, (x+y+z), is 22. This shows the importance of algebra in explaining such simple yet interesting mathematical phenomenon.

Do you know any such interesting property of any number?

5 Quick Shortcuts to Solve Time and Distance Problems

Time, Speed and Distance (TSD) is one of the most frequently occurring topics in quantitative aptitude section of many competitive exams. Here I’m putting together 5 short tricks which might come handy while answering various type of TSD questions. Let us start with some absolutely basic concepts –

Shortcut #1: Basic Concepts of Time, Speed and Distance

  • Speed =Distance/Time
  • Time =Distance/Speed
  • Distance = Speed x Time
  • To convert km/hr to m/sec
    • x km/hr =>x  * (5/18) m/sec
  • To convert m/sec to km/hr
    • y m/sec = y * (18/5) km/hr
  • If the ratios of the speeds of A and B is a :b, then the ratio of the times taken to cover the same distance is (1/a) :(1/b) or b:a

Shortcut #2: Finding out the Average Speed when Equal Distances are covered at Different Speeds

Lot of us make mistakes in calculation of average speed when the same distance is covered at different speeds.  We simply take the average of the given speeds. However, that gives absolutely wrong answer. So now get ready to find out what will give you the correct solution.

Theorem: If a certain distance is covered at x km/hr and the same distance is covered at y km/hr, then

Average speed =[2xy/(x+y)] km/hr

This is basically harmonic mean of the two speeds, i.e. 2/(1/x+1/y)

Example – If a car travels at 40 km/hr on a trip and at 60 km/hr on return trip. What is its average speed for the entire trip?

First thing we should be careful is we shouldn’t just average the 2 speeds. Overall average speed is not (S1+S2)/2.  From the above direct formula the answer will be

Solution: (2x40x60)/(40+60) = 48 km/hr

If we’ve to find the average of more than 2 speeds, average speed will be the harmonic mean of all such speeds

N / (1/a + 1/b + 1/c + 1/d)

Here N = 4, i.e. the number of variables (speeds in this case)

Shortcut #3: Finding out the Distance when Equal Distances Covered at Different Speeds and Total Journey Time is given

Theorem: A person goes to a destination at a speed of S1 km/hr and returns to his place at a speed of S2 km/hr, if he takes T hours, then

One way distance = Total time taken x (Product of two speeds) / (Addition of two speeds)

= T * {S1*S2/(S1+S2)}

Example: A boy goes to school at a speed of 3 km/hr and returns to village at 2 km/hr. If he takes 5 hours, what is the distance between the school and the village?

Solution: Distance between school and village is 5 * (3*2) / (3+2)=6 km

Shortcut #4.1: Finding out the Distance when Equal Distances Covered at Different Speeds

Theorem: A person goes to a destination at a speed of S1 km/hr and returns to his place at a speed of S2 km/hr, if he takes T1 and T2  hours respectively, then

One way distance = Total time taken x (Product of two speeds) / (Addition of two speeds)

= (T1 – T2) * {S1*S2/(S1-S2)}

Shortcut #4.2: Shortcut for “Early and Late to Office” Type Problems

The same shortcut used above can be used in these type of problems. Here you go –

TheoremA person covers a certain distance having an average speed of x km/hr, he is late by x1 hours but with a speed of y km/hr, he reaches his destination yhours earlier, hence

Required distance = Product of two speeds x Difference between arrival times/Difference of two speeds

Example: A man covers a certain distance between his house and office on bike. Having an average speed of 30 km/hr, he is late by 10 minutes. However, with a speed of 40 km/hr, he reaches his office 5 minutes earlier. Find the distance between his house and office?

In the above case, the required distance = (30×40)x0.25/(40-30) =  30 km

Please note: 10+5 = 15 minutes = 15/60 hours = 0.25 hours

Now since you know few time, speed and distance shortcuts, let’s try this time and speed puzzle

Shortcut #5: Finding Speed or Time Required after Crossing Each Other

Theorem: If two persons or trains A and B start their journey at the same time from two points P and Q towards each other and after crossing each other they take a and b hours in reaching Q and P respectively, then

Time speed and distance

Using this relationship you can find out the missing variables which can be either speed or time. Once these are known you can easily find the distance.

Example: Two, trains, one from Howrah to Patna and the other from Patna to Howrah, start simultaneously. After they meet, the trains reach their destinations after 9 hours and 16 hours respectively. The ratio of their speeds is:

Solution: using the above relationship, the ratio of their speed is √16/√9 = 4/3 or 4:3

Do you know more shortcuts for time, speed and distance problems?

Miles to Kilometers – Incredibly Simple Method of Conversion

Relationship between Miles and Kilometers

Mile is an English unit of measuring length or distance and it is equal to 5280 feet or 1760 yards  (for your information, 1 yard = 3 feet).
Whereas, Kilometer is the measure of length or distance in metric system, where it is equal to 1000 meters.

1 Mile = 1.609344 kilometers

Now if you have to convert few miles to kilometers, it will be a tedious task, isn’t it? Would you like to learn a method of doing this conversion mentally?

Here is simple yet effective method using which you will get an approximate value but in most cases that will suffice the purpose. So here you go.

You need to understand a very simple idea – the relationship between kilometers and miles can be closely represented by Fibonacci Series. Now what’s this Fibonacci Series?

For those of you who don’t know, we first need to understand the concept of Fibonacci sequence. First few numbers of the Fibonacci Series are the following. 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144…

If you’ve noticed in the above series the numbers are the sum of the previous two numbers in the series, starting with 0 and 1. As you see 0+1=1; 1+1=2; 1+2=3;…….34+55=89 and so on. Hence a series of numbers in which each number is the sum of preceding tweo numbers is called Fibonacci series or sequence. So to understand the calculation of conversion from miles to kilometers or the other way round,  let us use Fibonacci series

Conversion of Miles to Kilometres

Relationship of succeeding numbers in Fibonacci series closely matches the relationship between miles and km.

  • 3 miles = 5 km
  • 5 miles = 8 km
  • 8 miles = 13 km (12.8748 km to be exact)
  • 13 miles = 21 km (20.9215 km to be exact)
  • 21 miles = 34 km (33.796224 km to be exact)
  • 89 miles = 144 km (143.232 km to be exact)

However, what if you have to convert number (miles) which is not in Fibonacci series into km? Not to worry. There is a way out. You need to break out that number in to Fibonacci numbers. Covert the numbers as per the above scheme and add the resultant numbers to get the answer. For example; 85 = 55+21+8+1 After the conversion: 89+34+13+1 = 137 km (136.794 km to be exact)

Fibonacci sequence has many other wonderful properties and wide occurrence in nature, music, design, etc. You’ve just seen another application of Fibonacci sequence. The answers you get are close enough.

Why 1089 is a Wonderful Number?

This article is about a number that has some truly exceptional properties. That number is 1089

Most Amusing Property of 1089

Select a three digit number (where the units and hundreds digits are not the same) and follow these instructions:

Step 1: Choose any three-digit number (where the units and hundreds digits are not the same).

Let us randomly select the number 469

Step 2: Reverse the digits of the number you have selected

So reverse of 469 is 964

Step 3: Subtract the smaller number from the bigger one

964 – 469 = 495

Step 4: Once again reverse the digits of this difference

Reverse of 495 is 594

Step 5: Add the last two numbers

594+495 = 1089

This result will be the same for any 3-digit number chosen in step 1. Isn’t it astonishing that regardless of which number you select at the beginning, you will get 1089 as the result. Check out similar amusing property of 6174, which is also called Kaprekar Constant, named after Indian recreational mathematician D.R.Kaprekar.

Another Interesting property of 1089

Let’s look at the first nine multiples of 1,089:

1089 x 1 = 1089

1089 x 2 = 2178

1089 x 3 = 3267

1089 x 4 = 4356

1089 x 5 = 5445

1089 x 6 = 6534

1089 x 7 = 7623

1089 x 8 = 8712

1089 x 9 = 9801

I am sure you notice a pattern in the products. Look at the first and ninth products. They are the reverses of one another. The second and the eighth are also reverses of one another. And so the pattern continues, until the fifth product is the reverse of itself, known as a palindromic number

One More Unique Property of 1089

33^2 = 1089 = 65^2 – 56^2

The above representation is also unique among two digit numbers.

Do you agree that there is a particular beauty in the number 1089?

Mnemonics to Memorize Mathematical Concepts

Mnemonic (pronounced neh-MAHN-ik ) is a learning technique that aids information retention in our memory. It can be a rhyme, abbreviation or mental image that helps you in memorizing something which is otherwise difficult to commit to our memory.

In mathematics we come across lot of formulae, concepts and values that are torturous to memorize, especially when they are humungous in number. Let’s delve into few useful and handy mnemonics to memorize some very essential mathematical concepts / formula.

Mnemonic to memorize the value of pi (π)

Memorize till 7 decimal places: 3.1415926

May I have a large container of coffee?

mnemonic to memorize value of pi 7 place

Memorize till 10 decimal places: 3.1415926535

May I have a large container of coffee ready for today?

value of pi 10 places of decimals

Mnemonic to memorize the value of e (exponential function)

‘e’ (exponential function) is extensively used in calculus and in problems related to continuous growth.

If you wish to understand “e” in very simple terms, click here.

Value of e up to 15 places of decimal is 2.718281828459045

Value of e can be memorized by just breaking the value in chunks and memorizing it, like we do with mobile numbers. Here it goes –

mnemonic to memorize value of e exponential

2.7 – 1828 – 1828 – 45 – 90 -45

Mnemonics to Memorize Roman Numerals in Order

  • I = 1
  • V = 5
  • X = 10
  • L = 50
  • C = 100
  • D = 500
  • M = 1000

I Value Xylophones Like Children Drink Milk

OR

I Viewed Xerxes Loping Carelessly Down Mountains

Memorize Roman Numberals

Mnemonic to memorize the sequence to carry out the operations of arithmetic

There is a sequence in which we need to carry out arithmetical operations. This sequence can be memorized by the abreviation – BODMAS.

BODMAS

Position of division and multiplication can be interchanged as you know division is nothing but multiplication with inverse or reciprocal of the given number, example 13 ÷ 7 = 13 x 1/7

Position of addition and subtraction can be interchanged as subtraction is nothing but addition by changing the sign of the other number, example, 13 – 7 = 13 + (-7)

Mnemonic to memorize trigonometric Ratios

Often times it’s mind-numbing to memorize trigonometry ratios. Worst part is it keeps escaping out of our memories and we keep struggling.

Here I wrote about a way to easily remember trigonometry ratios

Some People Have, Curly Brown Hair, Turned Permanently Black

Memory trick for trignometric ratios

Do you think these mnemonics are interesting and helpful?
Do you have any such mnemonic to share?

Learn to Solve Simultaneous Equations Mentally

Do you struggle with solving simultaneous equations?

By simultaneous I mean equations with multiple unknown variables. Generally the number of equations given will be equal to the number of equations.

Let’s take an example,
3x + 4y = 18
5x + 7y = 31

Methods Taught at Schools

In our schools, we are taught to solve for x by equating the co-efficient of y by multiplying both the equations by some constants in such a manner that you get the same resultant value and then subtracting one equation from the other.

For instance in this case, to find the value of x, we will multiply first equation by 7 and second equation by 4 and get 28 in both the cases. This is done so that we get a zero on subtracting.
3x + 4y = 18 …………………(i) x 7
5x + 7y = 31 …………………(ii) x 4

We get 2 equations, where co-efficient of y is same
21x + 28y = 126
20x + 28y = 124

Subtracting second equation obtained above from the first one we get
(21x – 20x) + (28y – 28y) = 2
Hence, x = 2

Plugging the value of x in equation (i) we get y = 3

Problems with the Above Methods

  • This method can become quite laborious, especially when the co-efficients of the unknowns are such that they have to be multiplied by large numbers to make them equal to eliminate one of the unknown by adding or subtracting as the case may be.
  • The above calculations become cumbersome, when the co-efficient(s) are large prime numbers.
  • This method involves multiple steps where we need to do multiplication and addition/subtraction.
  • Also, there is no chance of using this method to solve the problem mentally as one has to keep track of the equations and various computations.

Solving Simultaneous Equations the Smarter Way

Our new method will give us the final answer in fractions, i.e. in Numerator and Denominator for both the variables: x and y.

First, we need to find the numerator of the value of x in the above case, take the simple following steps:
Step #1: Cross-multilply the coefficient of y in the first equation by the constant term (RHS) of the second equation
Step #2: Subtract from it the cross-product of the y coefficient in the second equation and the constant term (RHS) of the first equation.

Solving Simultaneous equation

So the numerator is 4×31 – 18×7 = 124-126 = -2.

Second, we need to find the denominator of the value of x:
Step #1: Cross-multiply the coefficient of y in the first equation by the coefficient of x in the second equation
Step #2: Subtract from it the cross-product of the y coefficient in the second equation and x coefficient in the first equation.

Solving linear equations

Hence the denominator is 4×5 – 7×3 = -1
Hence, the value of x = -2/-1 = 2

Now, let’s try with a simpler example,
x+2y = 8
3x + y = 9

Using the above method, in a single line calculation you can say,
x = (2×9 – 1×8)/(2×3 – 1×1)
x = 10/5 = 2
Therefore, y = 3

Isn’t this amazingly simpler? With some practice you can comfortably apply this technique to solve simultaneous equation mentally.

If you liked this method, you must explore another Vedic Mathematics trick of solving a special class of simultaneous equations in seconds.

Do you find simultaneous linear equations difficult to solve? Do you think you can start using above method in solving equations?

Understanding Exponential Function – “e” from Layman’s Perspective

In our journey of learning mathematics, we must have across a mathematical constant called “e”. It is approximately equal to 2.71828.

Unfortunately, most of us memorized the value and its usage without bothering about the concept behind it. That’s because mostly mathematics is taught in a way, where we try to explain concepts by their technical features without really explaining what it is and why it is used in the first place. Let us explore very basic insights about this so called “e”.

The natural exponential “e” explained in simple terms

In simple terms, “e” is manifestation of continuous growth in any naturally and continuously growing phenomenon.

Now what is this continuous growth and how do we deal it mathematically? To answer this question, we need to start with the other type of growth which we understand more easily, i.e. discrete growth.

Discrete Growth

Discrete growth can be visualized as step by step growth at the end of each period. If something doubles after every period; that is, at 100% growth per period

  • 1 becomes 2 at the end of 1st period;
  • 2 become 4 at the end of 2nd period;
  • 4 become 8 at the end of 3rd period; and so on…

This is basically 2^n where n is the number of periods. Remember, in the case of discrete growth nothing happens till we don’t reach the end of the period, then suddenly it doubles. It’s more like growth in step by step fashion depicted in the diagram below.  If you’re just interested in knowing how much time it takes to double your money at certain rate of interest, you will love the Rule of 72.

Discrete Growth vs Continuous Growth

In real life, the growth of anything like money, population, bacteria, plants, etc. is more smooth i.e. it is continuous process rather than discrete. 

Given the following formula for compounding, we get discrete results –

A = P (1 + r/n) ^ nt

  • A = Is the future value of investment including interest
  • P = Principal initial investment or the present value of the investment
  • r = yearly rate
  • n = number of compounding periods
  • t = no. of years

Assuming, the initial amount is Re.1, what happens when the yearly rate of growth is 100% and t is 1 year?

The above compounding formula gives us,

compunding computation

Calculation credit: Purplemath

Continuous Growth

Now, think how we can, mathematically shift from discrete stairs like growth curve to continuous smooth growth curve?

Compound_Interest_with_Varying_Frequencies

From the above table and graph, you know intuitively that higher the value of n, smoother the curve will become. So continuous growth is possible when the growth happens over infinitely small periods of time.  Thus to make it continuous n is assumed to be “x” which tends to infinity.

e_definition

Thus, you have got a new number “e” which denotes continuous growth over 1 period in which growth is taken as 100%.

Different Growth Rates and Time Periods

Now the question in your mind must be what if the growth rate is not 100% per period but something else. Fair question.

Remember, “e” by itself expresses the continuous growth in something at the rate of 100% over 1 time period, say 1 year. At 100% growth r becomes 1 (as you know 100% = 1) and for single time period t = 1.

Essentially, continuous compounding is expressed as e^x or it can be further expanded as ert where r is the rate of interest and t is time period.

The number “e” is about continuous growth. For all practical purposes, we can use e^x, where x allows us to merge time and rate. 5 years at 100% growth is the same as 1 year at 500% growth, when continuously compounded. Intuitively, e^x means, how much growth do I get after after x units of time and 100% continuous growth.

So if rate is 20% and time is 3 years, the resultant growth at continuous compounding can be captured using e^rt = e^(0.2*3) = e^0.6. That’s like saying 100% growth rate for 0.6 time years

Hope now “e” has started making more sense to you. Yes or No, please let me know in comments below.